# Scheduling problem

1. Jun 26, 2005

### Jimmy Snyder

The probability that project a will be complete at time t after it begins is given as:

$$$p_a(t) = \left\{ \begin{array}{ll} 0 & \mbox{if t \leq t_1} \\ \frac {(t - t_1)}{t_2-t_1} & \mbox{if t_1 \leq t \leq t_2} \\ 1 & \mbox{if t_2 \leq t} \right.$$$

In other words, the project requires at least $t_1$ to complete and will take no more than $t_2$. The probability of completion at any time between $t_1$ and $t_2$ is a linear function of t.

The probability that project b will be complete at time t after it begins is given as:
$$$p_b(t) = \left\{ \begin{array}{ll} 0 & \mbox{if t \leq t_3} \\ \frac {(t - t_3)}{t_4-t_3} & \mbox{if t_3 \leq t \leq t_4} \\ 1 & \mbox{if t_4 \leq t} \right.$$$

Project b will begin the moment project a is complete. Given t, what is the probability $p_{ab}(t)$ that both projects will be complete at time t. Obviously,
$$$p_{ab}(t) = \left\{ \begin{array}{ll} 0 & \mbox{if t \leq t_1 + t_3} \\ 1 & \mbox{if t_2 + t_4 \leq t} \right.$$$.

What is $p_{ab}(t)$ for $t_1 + t_3 \leq t \leq t_2 + t_4$?